1975 IMO Problems/Problem 1

Problem

Let $x_i, y_i$ $(i=1,2,\cdots,n)$ be real numbers such that $x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.$ Prove that, if $z_1, z_2,\cdots, z_n$ is any permutation of $y_1, y_2, \cdots, y_n,$ then $\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.$

Solution

We can expand and simplify the inequality a bit, and using the fact that $z$ is a permutation of $y$ , we can cancel some terms. $\sum^n_{i=1}x_i^2 + \sum^n_{i=1}y_i^2 - 2\sum^n_{i=1}x_iy_i \leq \sum^n_{i=1}x_i^2 + \sum^n_{i=1}z_i^2 - 2\sum^n_{i=1}x_iz_i$ $\sum^n_{i=1}x_iy_i \geq \sum^n_{i=1}x_iz_i$ Consider the pairing $x_1 \rightarrow y_1$ , $x_2 \rightarrow y_2$ , ... $x_n \rightarrow y_n$ . By switching around some of the $y$ values, we have obtained the pairing $x_1 \rightarrow z_1$ , $x_2 \rightarrow z_2$ , ... $x_n \rightarrow z_n$ . Suppose that we switch around two $y$ -values, $y_m$ and $y_n$ , such that $y_m > y_n$ . If $x_m > x_n$ , call this a type 1 move. Otherwise, call this a type 2 move.

Type 2 moves only increase the sum of the products of the pairs. The sum is increased by $x_n \cdot y_m + x_m \cdot y_n - x_m \cdot y_m - x_n \cdot y_n$ . This is equivalent to $(x_n - x_m)(y_m - y_n)$ , which is clearly nonnegative since $y_m > y_n$ and $x_n > x_m$ .

We will now consider switching from the $x$ - $z$ pairing back to the $x$ - $y$ pairing. We will prove that from any pairing of $x$ and $z$ values, you can use just type 2 moves to navigate back to the pairing of $x$ and $y$ values, which will complete the proof.

Suppose that $x_a$ is the biggest $x$ -value that is not paired with its $y$ -value in the $x$ and $z$ pairing. Then, switch this $y$ -value with the $y$ -value currently paired with $x_a$ . This is obviously a type 2 move. Continue this process until you reach back to the $x$ - $y$ pairing. All moves are type 2 moves, so the proof is complete.

~mathboy100

Solution 2

We can rewrite the summation as $\sum^n_{i=1} x_i^2 + \sum^n_{i=1} y_i^2 - \sum^n_{i=1}x_iy_i \le \sum^n_{i=1} x_i^2 + \sum^n_{i=1} z_i^2 - \sum^n_{i=1}x_iz_i.$ Since $\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2$ , the above inequality is equivalent to $\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i.$ We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of $\sum^n_{i=1}x_iz_i$ . Obviously, if $x_1 = x_2 = \ldots = x_n$ or $y_1 = y_2 = \ldots = y_n$ , the inequality is true. Now, assume, for contradiction, that neither of those conditions are true and that there exists some order of $z_i$ s that are not ordered in the form $z_1 \ge z_2 \ge \ldots \ge z_n$ such that $\sum^n_{i=1}x_iz_i$ is at a maximum out of all possible permutations and is greater than the sum $\sum^n_{i=1}x_iy_i$ . This necessarily means that in the sum $\sum^n_{i=1}x_iz_i$ there exists two terms $x_pz_m$ and $x_qz_n$ such that $x_p > x_q$ and $z_m < z_n$ . Notice that $x_pz_n + x_qz_m - (x_pz_m + x_qz_n) = (x_p-x_q)(z_n-z_m) > 0,$ which means if we make the terms $x_pz_n$ and $x_qz_m$ instead of the original $x_pz_m$ and $x_qz_n$ , we can achieve a higher sum. However, this is impossible, since we assumed we had the highest sum. Thus, the inequality $\sum^n_{i=1}x_iy_i \ge \sum^n_{i=1}x_iz_i$ is proved, which is equivalent to what we wanted to prove. $\[\]$ ~Imajinary

Remark

It is only the most common way of rearrangement inequality after expanding and subtracting same terms.~bluesoul

1975 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1975 IMO Problems/Problem 1

Contents

Problem

Solution

Solution 2

Remark

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